Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $p = \dfrac{6k - 12}{10k + 10} \times \dfrac{-5k^2 + 45k}{k^2 - 11k + 18} $
Answer: First factor the quadratic. $p = \dfrac{6k - 12}{10k + 10} \times \dfrac{-5k^2 + 45k}{(k - 2)(k - 9)} $ Then factor out any other terms. $p = \dfrac{6(k - 2)}{10(k + 1)} \times \dfrac{-5k(k - 9)}{(k - 2)(k - 9)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ 6(k - 2) \times -5k(k - 9) } { 10(k + 1) \times (k - 2)(k - 9) } $ $p = \dfrac{ -30k(k - 2)(k - 9)}{ 10(k + 1)(k - 2)(k - 9)} $ Notice that $(k - 9)$ and $(k - 2)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -30k\cancel{(k - 2)}(k - 9)}{ 10(k + 1)\cancel{(k - 2)}(k - 9)} $ We are dividing by $k - 2$ , so $k - 2 \neq 0$ Therefore, $k \neq 2$ $p = \dfrac{ -30k\cancel{(k - 2)}\cancel{(k - 9)}}{ 10(k + 1)\cancel{(k - 2)}\cancel{(k - 9)}} $ We are dividing by $k - 9$ , so $k - 9 \neq 0$ Therefore, $k \neq 9$ $p = \dfrac{-30k}{10(k + 1)} $ $p = \dfrac{-3k}{k + 1} ; \space k \neq 2 ; \space k \neq 9 $